math 3 week 4 lecture slides [compatibility mode]

8/2/2019 Math 3 Week 4 Lecture Slides [Compatibility Mode]

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Linear Algebra

Review of matrices

ystems o near a ge ra c equat ons =Finding solution(s) by row operations

Inverse of a s uare matrix A

Finding inverse by row operations

Determinant of a square matrix ACalculating determinant by row operations

In between, we will look at vector space and linearly independentvectors

a r x e genpro em

Diagonalisation problem


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Review on matrices

An MNmatrix is an array of MNnumbers enclosed within apair of brackets and arranged in Mrows and Ncolumns.

1

2 1 3 6 9

Examples:

( 3 ) 21 2 2 3 92

5 6

The numbers making up a matrix are referred to as elementsof the matrix.

,columns.

We use bold capital letters such as A, B, P and Q to denote. . .

1 2 3

4 5 6

A


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Notation

Let A be an MNmatrix.

enote t e e ement n t e -t row an -t

column of A by aij. Then we can write:

11 12 1

21 22 2

N

N

a a a

a a a

ij

a a a


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Equality of matrices

ij ij .

, , , ij ij.

E.g. 3a b c c

, 3, ,a b c c c d a b b

on o ma r ces

If A+B = C then C=(cij) is MNand cij= aij+ bij.

1 2 3 7 8 9 1 7 2 8 3 9

E.g.

4 5 10 11 12 4 10 5 11 12


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Multiplication of a number to a matrix

If A = (aij) and cis a number then cA = (caij) andcA has the same order as A.

E.g. 1 2 2 423 4 6 8

We write (1)A as A. So BA = B + (A).

E.g.2 3 1 2 2 1 3 2

4 4 3 2 4 3 4 2


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The trans ose of a matrix A is denoted b AT

.1 2 3 1 4 7

7 8 9 3 6 9

= T

1 1

3 0 6 3 0 6 T

A A

5 6 9 5 6 9


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Product of matrices

Let A=(aij) and B=(bij) be MNand PQmatricesrespectively.

= , we can orm t e pro uct matr x .

If AB is denoted b C= c then C is MQand the

element ckp is calculated using the k-th row of A and p-th

column of B as follows.

kc

1

2

1 2

p

p

k k kN

ba a a

1 1 2 2k p k p kN Np

a b a b a b

A typo here in the 238page document (p56)

Npb Nk n n ja b

What is the condition for forming the product BA?

What is the order ofBA if it can be formed?


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Example:

2 4

5 1 2 23 4

3 3 1 2

P Q

3 2

We can form PQ but not QP.

1 2 11 7 4 6

5 1 2 2

2 4

3 2 3 4

3 3 1 25 6 43 23 16 22

, ,BA can be formed, AB may or may not be equal to BA.


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Identit matrices

An NNmatrix (cij) such that c11 = c22 = c33== cNN=1 and cij= 0for inot equal tojis called an identity matrix.

1 0 0 01 0 0

Examples:

0 1 00 1 0 0 1 0

0 0 1 0 0 0 1

Whats special about identity matrices?

.If the product IA can be formed then IA = A.

, .


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Lets start on linear algebra now

A good starting point is to look at asystem of linear algebraic equations.

Whats a linear algebraic equation?

Many simultaneous linear algebraic equationsform a system.

How can we solve a system of linearalgebraic equations?


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An example of a linear algebraic equation in

one unknown x is: 2 1 0x

Thesolution of the above linear algebraicequation is:

2

x

is an equation of the form:

If we let y=0 then x= 3. So (x,y)=(3,0) is a solution of the above equation.


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near a ge ra c equa on n un nowns x1, x2,

,xNand xNis an equation of the form:

1 1 2 2...

N N N c x c x c x d

Why bother


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Many problems in engineering and

p ys ca sc ences are ormu a e nterms of a systemof linear algebraicequa ons.

This tem eratureprofile in the human

eye was obtained from

method by solving asystem of 470 linear

algebraic equations in470 unknowns.


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Heres a system ofNlinear algebraicequations inNunknowns.

11 1 12 2 13 3 1 1

21 1 22 2 23 3 2 2

...

...

N N

N Na x a x a x a x b

1 1 2 2 3 3... N N N NN N N a x a x a x a x

ij is the (constant) coefficient of the unknown xjin the i-th equation.

-i

.


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The system can be written in matrixform AX =B, where:11 12 1

...N

a a a Example:

12 22 2...

Na a a

A

2 3 1 0x y 1 2 N N NN

1

2

x

x

2 3 10x

Nx

1 1 0y

1

2

b

b B

2 3 10x y

Nb


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Solutions of a linear al ebraic e uationConsider a single linear algebraic equation

1 1 2 2...

N N N

1 1x is said to be a solution of the above linear

2 2

x

a ge ra c equat on t e o t e equat onequals its RHS when we replace x1, x2, , xN

by 1, 2, , Nrespectively.

Example: We can find many other solutions easily.

2 0 x y z 1x 2x

3

y

z

s a so u on u s no .

3z


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Solution of a system of linearalgebraic equations

1 1x

2 2x

If is a solution ofN Nx

each and ever linear al ebraic

equation in the system then it is.


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It is possible that a system of linearalgebraic equations has no solution.

Example:

5

x y

x

.


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has more than one solutions.

xamp e:

1 0x y

2 2 2 0x y 10x y

The system really contains only one linear algebraicequation in 2 unknowns!

.


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If a consistent system of linear algebraice uations has onl one solution we sa

the system has a unique solution.

To summarise, a system of linear algebraic

equations can either be consistent or inconsistent.If it is consistent, it can have either a uniquesolution or infinitely many solutions.


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Given a system of linear algebraicequations AX = B, how do we know

whether it is consistent or not?

,its solutions?


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Reduce the system AX = B to asimpler but equivalent systemUX = C.

AX = B and UX = C are equivalent if they haveexact y t e same so ut on s .

If we can work out the solution(s) of UX = C,= .

If the square matrix U is an upper triangular

, =enough for us to work out its solution(s) (if any).


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a s an upper r angu ar ma r x

1 4 5 6 E.g.

0 1 6 8

0 0 2 0


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To solve AX = B, reduce it to an

equ va en sys em = , w ereU is an upper triangular matrix.

How can this be done?

Write AX = B in tableau form A | B.

2 3 4 6 x y z 2 3 4 6

E.g.E.g.

3 5 2 7

10 5 9

x y z

x y z

1 10 5 9

Use legitimate row operations in a systematic


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There are 2 types of legitimate rowoperations.

R R Interchange i-th andj-th rows.

i i j R R R Use rowj to change row i tobecome Ri +R .Important. The constant is not allowed to be zero.

Why? WhyR1 R3 (say) is not allowed?

A simple rule to observe.

In changing a tableau by the second type of rowoperation, keep one row fixed. Use the fixed row


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Example:

2 3 2 x y z w

2 4 1 x y z

8 y z w

1 1 2 3 2

1 1 1 1 5

0 2 2 1 R R R 0 1 -2 3

2 1 4 0 1 0 3 3 12 R R R -3 8 -6 -3


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1 1 2 3 2 0 0 1 2 3 0 1 1 1 8 2 4R R

0 1 1 1 8

0 0 1 2 3

1 1 2 3 2

0 1 1 1 8

0 1 1 1 8

0 0 11 3 21

0 0 0

0 0 1 2 30 3 3 23 R R R 11 -3 21

0 -19 12

4 4 311 R R R


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x y z w

1 1 2 3 2

0 1 1 1 8

8z w

2 3 2 x y z w

0 0 11 3 21 11 3 21 (21 3 ) /11 33/19 z w z w w w

e on y so u on o e sys em s

9/19x

131/19

33/19

y

z

12/19w


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Linear Algebra

- Review of matrices

ystems o near a ge ra c equat ons =Finding solution(s) by row operations

Inverse of a s uare matrix A

Finding inverse by row operations

Determinant of a square matrix ACalculating determinant by row operations

a r x e genpro em

Diagonalisation problem


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Review

We were looking at solving a systemf lin r l r i i n .

A X = B a system of linearAX Blegitimate row i jR R

UX C( 0)i i j R R R

UX = C equivalent systemupper triangular matrix

xamp e on p


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Example

2 3 6 x y z 1 2 3 6

8 9 10 0

x y z

x y z

8 9 1 0 0

1 2 3 6

0

2 2 1

3 3 18 R R R 7 14 48


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1 2 3 6

0 4 8 22

0 7 14 48

1 2 3 6

0 4 8 22 0 0

3 3 24 7 R R R 0 38

The last row gives a nonsensical statement!

.


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Example x y z

2 3 8

5 6 7 24

x y z

x

1 2 3 8

0 4 8 16

8 9 10 36 x y z 0 0 0 0

There is nothing wrong with the last row. It tells that

there are only 2 independent equations. The system.

Let z= s(sis any arbitrary real number). x s

y s y s 2(4 2 ) 3 8 x s s x s

4 2y s

z s

Read note at the bottom of p70.


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Example x y z w

1 2 3 2 1

0 1 0 0 2

2 3 2 1

2 5 6 4 0

x y z w

x y z w

0 0 0 0 0

0 0 0 0 0

3 7 9 6 1

3 3 2 1

x y z w

x y z w

There are only two independent equations here.

We can let two of the unknowns be any valuesbut second row in the final tableau tells us thaty= 2.

1 22( 2) 3 2 1 x t t

e = 1 an w= 2.


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The Bi Picturoriginal system

le itimate rowoperations

UX C simpler but equivalentsystem


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A homogeneoussystem of linear

a11x1 + a12x2 + +a1(N1)xN 1 +a1NxN= 0

a21x1 + a22x2 + +a2(N 1)xN 1 +a2NxN= 0=

aN1x1 + aN2x2 + +a N 1 xN 1 +a NxN= 0

In matrix form, it can be written as AX = 0

where 0 = .0

0


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A homogeneous system AX=0 is

always consistent.

1

2 0 x is a solution, no matter what A is.

0N

x

trivial solution

A | 0 U | 0

Depending on what A is, the homogeneous systemAX=0 has either only one (unique) solutiongiven byX=0 or infinitely many solutions(one of which is X=0).


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n a so ut ons o t e omogeneous system

2 3 0 x y z 1 2 3 0

5 6 7 0

8 9 10 0

x y z

z z z

8 9 1 0 0

1 2 3 0

0 5 R R R -4 -8 00

3 3 18 R R R -7 -14 0

2y t

0 4 8 0 x t

infinitely many

0 3 3 2so u ons


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A | B U | C

We can make the following general observations:

a t e agona e ements o are not zero,

the system AX=B has only one solution.For a homo eneous s stem this means X=0

is the only solution.)

,the system AX=B has either no solution orinfinitely many solutions.

or a omogeneous system, t s means t atthere are infinitely many solutions, one ofwhich is X=0.


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Review

Solving a system of linear algebraici n :

A X = B a system of linearAX Blegitimate row

i jR R

UX C( 0)i i j R R R

UX = C equivalent systemupper triangular matrix


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Example: 5

3

1 3

,4

and .

0

1

,

0 1 3

If possible, finda, band csuch that

3 0 3ba

-3 =-4

1 0 10 1 3

+ b + ca

++ 3

cb c

= a

If possible, solve: a + 3b+0c= 5a + 0b+ c= 3

0a + b+3c= 4


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Write the equations in tableau form:

1 3 0 5

a b c1 3 0 5

1 0 1 3

0 1 3 4

0 1 3 4

2 2 1

1 3 0 5

3b+(2) = 8 b= 2

0 3 1

0 0

8 3 R R R 10 20

a+3(2) = 5 a = 1


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5

1 3

,4

and

0

1

,

0 1 3

Specifically, we find that:

3 ( 1) 1 2 0 ( 2) 1

4 0 1 3

More examples: Problems 5 and 6 on page 86

Li l i d d t t


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Linearly independent vectors

1, 2, , P1 P ,N elements.

independent vectors if we cannot find any oneof these vectors to be a linear combination ofthe other vectors.

How do we check if w1, w2, , wP1 and wParelinearly independent

Form the homogeneous system

c1 w1 + c2 w2 + + cP1 wP1 + cPwP= 0If c1 = c2 = = cP1 = cP = 0 is the only solutionof the system then the vectors are linearly

independent. Why does this work?

H d h k if d


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How do we check if w1, w2, , wP1 and wPare

Form the homogeneous system

1 1 2 2 P1 1

If c1 = c2 = = cP1 = cP = 0 is the only solution

independent. Why does this work?

, 1 we can write:

32 Pcc c

1 2 3

1 1 1

P

c c c

That is, we can express w1 as a linear combination of w2 , w2 , ,wP1 an wP , ence e vec ors are no near y n epen en .


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p77

Are

1 2 1

1 , 1 , 0

linearly independent?

1 1 2 0c1 c2 c3

1 2 30 1 1 0

1 1 1 0

c c c

1 1 2 00 1 1 0

-1 -1 -1 0

Is c1 = c2 = c3 = 0 the onlysolution?

0 1 1 0Vectors are linearly independent.

1 2 1 What about ?


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1 , 2 , 2 What about ?

2 4 3

1 2 31 2 2 02 4 3 0

c c c

1 2 2 02 4 3 0

1 2 1 0

0 0 1 0

c1=c2=c3=0 is not the only solution.

A possible non-trivial solution is:

1 = , 2 = 3 = .

1 2 1 0 2 1

Vectors are not linearlyinde endent.

2 4 3 0 4 2

math 3 week 4 lecture slides [compatibility mode]

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